= Solution
Use a nonzero complex unital <Banach algebra> $B$ with submultiplicative <norm>, and write its <unit> as $e$. The complex <scalar field> is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.
If $\|a\|<1$, <completeness> makes the <Neumann series> $\sum_{j=0}^\infty a^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\|b^{-1}h\|<1$. This proves that the invertible elements form an <open set> and that inversion is locally norm-continuous.
Define the <Banach algebra spectrum> and the <resolvent of an element> by
$$
\sigma(x)=\{\lambda\in\mathbb C:\lambda e-x\text{ is not invertible}\},
\qquad R(\lambda)=(\lambda e-x)^{-1}\quad(\lambda\notin\sigma(x)).
$$
For $|\lambda|>\|x\|$, the <Neumann series> gives
$$
\boxed{R(\lambda)=\sum_{j=0}^\infty\frac{x^j}{\lambda^{j+1}},\qquad x^0=e.}
$$
Thus $\sigma(x)$ is bounded by $\|x\|$ and is <closed>, because invertibility is open. Also $R(\lambda)\to0$ in <norm> at infinity; for example its <norm> is at most $\|e\|/(|\lambda|-\|x\|)$, since $\|e\|\geq1$ in a nonzero submultiplicative unital algebra. The standard normalization $\|e\|=1$ is not needed.
For points where the <Banach algebra resolvent> exists $\lambda,\mu$, multiply out the inverses to get the <resolvent identity>
$$
\boxed{R(\lambda)-R(\mu)=(\mu-\lambda)R(\lambda)R(\mu).}
$$
For $\lambda_0$ outside the <Banach algebra spectrum>, put $R_0=R(\lambda_0)$. Locally,
$$
R(\lambda_0+h)=R_0(e+hR_0)^{-1}
=\sum_{j=0}^\infty(-h)^jR_0^{j+1},\qquad |h|\|R_0\|<1.
$$
This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a <Banach-space-valued holomorphic function>, establishing complex differentiability in the Banach-space <norm> and gives $R'=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every <continuous> complex <linear functional> $\phi$ makes $\phi(R(\lambda))$ a scalar <holomorphic function>.
The <Banach algebra spectrum> is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local <norm> continuity and decay at infinity make it globally bounded. For every <continuous linear functional> $\phi$, the scalar <entire function> $\phi(R(\lambda))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar <Cauchy estimate> for its <derivative> on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the <Hahn-Banach theorem>. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its <norm>, and complexifying it by $\phi(v)=\ell(v)-i\ell(iv)$ gives a bounded complex <linear functional> nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved \b[the <Banach algebra spectrum> is a nonempty compact subset of the <complex plane>].
We next establish the <spectral radius formula>. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\leq r<m$ to obtain
$$
\|x^n\|\leq\|x^m\|^q\max_{0\leq r<m}\|x^r\|.
$$
Taking roots gives $\limsup_n\|x^n\|^{1/n}\leq\|x^m\|^{1/m}$. Taking the <infimum> in $m$, and using that every root <norm> is at least that <infimum>, proves existence of
$$
\rho_0=\lim_{n\to\infty}\|x^n\|^{1/n}
=\inf_{m\geq1}\|x^m\|^{1/m}.
$$
For $|\lambda|>\rho_0$, the <root test> makes $\sum x^j/\lambda^{j+1}$ converge, even when $|\lambda|\leq\|x\|$. The same telescoping <multiplication> proves it is a two-sided inverse. Hence
$$
r_\sigma:=\max_{\lambda\in\sigma(x)}|\lambda|\leq\rho_0.
$$
For the reverse bound fix $R>r_\sigma$. If $R_1>\max(R,\|x\|)$, integrate the uniformly convergent exterior series term by term on $|\lambda|=R_1$, obtaining
$$
x^n=\frac1{2\pi i}\int_{|\lambda|=R_1}\lambda^nR(\lambda)\,d\lambda.
$$
The same integral equals the one on $|\lambda|=R$: applying an arbitrary <continuous linear functional> reduces <contour deformation> through the <Banach algebra spectrum>-free annulus to the scalar <Cauchy theorem>, and separation of points restores equality in $B$. This proves the <resolvent Cauchy coefficient formula> rather than assuming vector-valued analytic calculus. Using the <norm> bound for the allowed vector integral gives
$$
\|x^n\|\leq R^{n+1}\max_{|\lambda|=R}\|R(\lambda)\|.
$$
Taking $n$th roots yields $\rho_0\leq R$, and letting $R\downarrow r_\sigma$ finishes the proof:
$$
\boxed{\rho(x)=\lim_{n\to\infty}\|x^n\|^{1/n}
=\max_{\lambda\in\sigma(x)}|\lambda|
=\sup\{|\lambda|:\lambda e-x\text{ is not invertible}\}.}
$$
For the first example take the <commutative algebra> of <dual numbers> $B=\mathbb C[\varepsilon]/(\varepsilon^2)$ with <norm> $\|a+b\varepsilon\|=|a|+|b|$. <Completeness> follows from finite dimension, and the product estimate makes the <norm> submultiplicative. For $x=\varepsilon$, $x\ne0$ but $x^2=0$, so $\rho(x)=0$. Directly, $\lambda e-\varepsilon$ has inverse $\lambda^{-1}e+\lambda^{-2}\varepsilon$ for $\lambda\ne0$, and its <Banach algebra spectrum> is $\{0\}$. For the second example take $B=\mathbb C$ with its absolute-value <norm> and $x=1$. Its <Banach algebra spectrum> is $\{1\}$, giving \b[$\rho(x)=\|x\|=1$].
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